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2 years ago

Good afternoon help me 5 questions is 50 points and answer them all

Mathematics

1 answer:

Paraphin [41]2 years ago

5 0

The first one is the last answer because you subtract 18 and when you divide you change the inequality symbol

the second one is the last answer for the same reason

the 3rd one is x<-1.25

the 4th one is x<8/15

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It is 6 kilometers from Linda's house to the nearest mailbox. How far is it in meters?

Zanzabum

Answer:

6000 meters

Step-by-step explanation:

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1 year ago

What is y≤-3x-7 and y≥3/4x+8

miskamm [114]

Answer: (-4,5)

Step-by-step explanation:

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2 years ago

Which is a true statement?

julsineya [31]

B. <span>–4.37 < –1.63 is the true statement. -4.37 is less than -1.63
</span>
A. <span>–4.28 > –3.56 * should be less than
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In dealing with negative numbers, the greater the negative number is, the lesser its value. 4.37 is a greater number than 1.63 in its positive form, however, its negative form give 4.37 a lesser value than 1.63 in its negative form.

7 0

3 years ago

Read 2 more answers

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$